TMJC 9758 2025 Prelim P2 Solutions

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 1 of 17 2025 H2 MATH (9758/02) JC 2 PRELIMINARY EXAMINATION Qn Solution 1 Equations and Inequalities (a) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 2 5 12 5 3 5 102 5 3 5 2 5 3 02 5 3 5 2 5 3 02 5 3 2 4 2 02 5 3 2 2 1 02 5 3 21 10, , 33 2 1 2 x xx x xx x x x xx x x x xx xx xx xx xx x xxxx + − + + + −− + + + − − + + − + + + + − − − + + −+ − + + −+ − + + −  − −+ 1 or 32xx−  (b) 2 cos 5 12cos 5cos 3 x xx + − + + Replace x by cos x : 1cos or cos 3 (Rej. 1 cos 1)2x x x−  −   2 3 x    0 x y 1 + + 3 − −

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 2 of 17 Qn Solution 2 Maclaurin Series (a) Since the curve ( ) 2f a b c  = + + passes through ( )0,1 , 1c= Since the curve ( ) 2f a b c  = + + passes through 31,32 − , 2 3 3 1 1 3 1 13 3 2 3 3 2a b a b   − + − + =  − =−          …….(1) ( )f ' 2 ab=+ Since the turning point is at 31,32 − , ( )f ' 0 = ( ) 320 3 23 03 23 23 . ab ab ba − + =  += =   − Sub (2) into (1): 1 3 1 3 3 2 1 3 2 3 1 3 3 3 2 1 1 3 3 2 2 ab aa aa − =− − =− − =−  = 2 3 2 3 3 33 3 2ba = = =  3 , 3, 12a b c= = = (b) ( )( ) 2 2 2 2 cos 6BC AB AC AB AC = + − +  ( ) ( )( ) 221 3 2 1 3 cos cos sin sin 66 314 2 3 cos sin22 4 3cos 3 sin    = + − −  = − −  = − +

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 3 of 17 ( ) 2 2 2 when is small ) 4 3 1 3 2! 34 3 3 2 313 2 f (Shown        − − + = − + + = + + = (c) ( ) 2 1 22 2 22 22 2 313 2 313 2 11 11 3 3 221 3 32 2 2! 2 3 3 113 2 4 8 331 (Shown)28 BC            + +  + +  −     + + + +        + + − = + +

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 4 of 17 Qn Solution 3 Graphing Techniques, Transformation of Curves (a) (b) x y O x y O

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 5 of 17 Qn Solution 4 Differential Equations (a) 22 2 dd 0.05dd PP tt =−  Substituting d d Pv t= , we get 2d 0.05d v vt =− Using separable variables, 2 1 d 0.05 dvtv =− 1 0.05 'tcv− =− + where 'c is an arbitrary constant 1 0.05 ' 20 v tc tC = − = + where 20 'Cc=− is a constant (shown) (b) 20v tC= + Given 0 and 10Pv== when 0,t= 2010 2 CC=  = 20 2v t= + d 20 d2 P tt= + 20ln 2P t D = + + where D is a constant when 0,t= 0,P= 0 20ln 2 20ln 2 D D  = + =− 20ln 2 20ln 2 220ln since 02 Pt t t = + − +=  (c) When 30 230, 20ln 55.5 2tP += = =  (3 s.f)

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 6 of 17 Qn Solution 5 Vectors (a) ( )  =   −  =  − = a b a c a b a c 0 a b c 0 Since a0 and ,bc OA= a is parallel to .CB=−bc (shown)−=b c a (b) Method 1: ( )( ) ( ) ( ) Heightsin 30 5Height 5sin 30 2 1Area of Trapezium Height2 1510 3 22 5 shown OA OA BC BC BC = = = =+ =+  = − =cb Method 2: ( )( ) ( ) ( ) ( ) ( ) 1Area of Trapezium height of trapezium2 110 2 1 5 3sin 3010 3 23 520 3 2 5 shownBC = + −  = + −   = + −  = + −  = − = a c b caa c b a cb cb cb Method 3: 1 1 1 15Area of triangle OAC sin 30 3 52 2 2 4 15 25Area of triangle ABC 10 44 =    =    = = − = ac 25 Area of triangle ABC 5 4 15Area of triangle OAC 3 4 == O A C B 5 3 Height

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 7 of 17 ( ) 1Area of triangle OAC 2 1Area of triangle ABC 2 Area of triangle ABC 5 5 shownArea of triangle OAC 3 3 h BC h BC BC BC =   =   = = =  = a a (c) 3 5k = (d) ( ) : , : , OC AB l l   = = + −  rc r a b a From (c), ( )3 5 3 5OA CB= =−a b c . At point D, ( )= + −c a b a ( ) ( ) ( ) : 3 5 3 3 3 3 5 5 5 5 3 3 3 3 5 5 5 5 3 2 3 3 55 5 5 3 5 ABl          = + − = − + −  = − + − +    = − + + −          = + + −    −    r a b a r b c b b c b b c bc b bc c At point D, 3320 3 5555   + + = + − b c b c By comparison, 2 32 05 3 5   + =− = and 2 3 3 3 3 5 5 2 33 55   = − = − −    =−  3 2OD =− c . Alternatively, Triangle DOA and triangle DCB are similar triangles.

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 8 of 17 5 53 3 15 5 7.5 7.5 5 3 2 DC DO CB OA xx xx x OD = + = += = =− =− c c D O A C B x 5 3 5

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 9 of 17 Qn Suggested Solutions 6 Probability (Venn diagram) (a) ( ) ( ) ( ) ( ) ( ) ( ) 6 P 3 0.4 0.12 0.42 0.1 P P [since and are independent] 0.3 P P P [since and are independent] 2 0. A B A B A B A C A C A C    =  = = = = = (b) Representing values into Venn diagram, (c) From Venn diagram, For minimum x, let ( )0 greatest P ' 0.294x A B C =    = For maximum x, let ( )0.28 least P ' 0.014x A B C =    = ( ) ( ) least P 0.014, greatest P 0.294. A B C A B C    =   = A B C 0.074 0.1 0.02 0.28 – x 0.106 0.294 – x 0.126 + x x

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 10 of 17 Qn Solution 7 Hypothesis Testing (a) Unbiased estimate of  is 6960 11660 xx n= = = ( ) ( ) 2 2 2 2 2 6 U 1 nbiased estimat 2 1 1 69608 246559 60 15105 59 25 6.0169 e of  is 25 x sx nn  =− −   =−   = = =  (b) Let X be the caffeine content per cup of signature coffee in mg. Let µ denote the population mean caffeine content per cup of signature coffee in mg. 0H : 120 = 1H : 120 Under 0H , since 60n= is large, by Central Limit Theorem, 256.0169N 120, 60X   approximately. Test Statistic: 120 256.0169 60 XZ −= Level of significance: 5% Reject H0 if value 0.05p− Using G.C, ( )value 0.0528 3 s.fp−= Since value 0.0528 0.05p− =  , we do not reject 0H and conclude that there is insufficient evidence, at the 5% level of significance, that the population mean caffeine content per cup of coffee is not 120 mg. Hence the coffee shop owner’s claim is supported by the data at the 5% level of significance. (c) Let Y be the caffeine content per cup of premium coffee in mg. Let Y denote the population mean caffeine content per cup of premium coffee in mg. 0H : 120Y = 1H : 120Y  Under 0H , ( )N 120,200Y , 200N 120,Y n   Test Statistic: 120 200 YZ n −= Level of significance: 2.5%